06-07 Ladder slides down the wall

$x^2 + y^2 = 20^2$

$2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0$

$x \dfrac{dx}{dt} + y \dfrac{dy}{dt} = 0$

$x \dfrac{dx}{dt} + y (-2) = 0$

$x \dfrac{dx}{dt} - 2y = 0$
 

when x = 16 ft
$16^2 + y^2 = 20^2$

$y = 12 \,\, \text{ ft }$
 

$16 \dfrac{dx}{dt} - 2(12) = 0$

$\dfrac{dx}{dt} = 1.5 \, \text{ ft/sec}$           answer

From the figure in Solution 6 above
$m = \dfrac{y}{x}$

$\dfrac{dm}{dt} = \dfrac{x \dfrac{dy}{dt} - y \dfrac{dx}{dt}}{x^2}$
 

From Problem 6:

x = 16 ft
y = 12 ft
dx/dt = 1.5 ft/sec
dy/dt = -2 ft/sec

 

$\dfrac{dm}{dt} = \dfrac{16(-2) - 12(1.5)}{16^2}$

$\dfrac{dm}{dt} = \dfrac{-50}{256}$

$\dfrac{dm}{dt} = -\dfrac{25}{128} \,\, \text{ per second } \,\, $

$\dfrac{dm}{dt} = \dfrac{25}{128} \, \text{ per second decreasing}$           answer